Log-cos-p-cos-10-p-sec-1-cos-1-10-p-10-p-Log-sec-Log-10-p-p-Log-10-

Question Number 117781 by hmh2k20 last updated on 13/Oct/20

Log (\cos β) = p \Rightarrow \cos β = 10^{p}

∴ \sec β = \frac{1}{\cos β} = \frac{1}{10^{p}} = 10^{- p}

∴ Log (\sec β) = Log 10^{- p}

= - p Log 10

= - p

Answered by $@y@m last updated on 13/Oct/20

K i n d l y u s e e d i t o r o p t i o n o f t h e a p p

f o r r o u g h w o r k .

J u s t t a p o n \equiv s i g n o n t o p l e f t s i d e

a n d t h e n t a p o n editor .

T h i s s e c t i o n i s m e a n t f o r m u t u a l

d i s c u s s i o n .

Commented by Tinku Tara last updated on 13/Oct/20

Latest updated app opens forum

be default and editor is opened

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Most of users are using the app

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few actually create user id . This

interface simplfies browsing for

majority of users .

Only a small set of users are actually

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Commented by $@y@m last updated on 13/Oct/20

Y o u a r e r i g h t .

O f f l i n e e d i t o r i s u s e d r a r e l y .

S o I a p p r e c i a t e n e w u p d a t e .

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